3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution

In this guide, you will get 3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution with the best time and space complexity. The solution to Count Submatrices With Equal Frequency of X and Y problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Count Submatrices With Equal Frequency of X and Y solution in C++
Count Submatrices With Equal Frequency of X and Y solution in Java
Count Submatrices With Equal Frequency of X and Y solution in Python
Additional Resources

3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution image

Problem Statement of Count Submatrices With Equal Frequency of X and Y

Given a 2D character matrix grid, where grid[i][j] is either ‘X’, ‘Y’, or ‘.’, return the number of submatrices that contain:

grid[0][0]
an equal frequency of ‘X’ and ‘Y’.
at least one ‘X’.

Example 1:

Input: grid = [[“X”,”Y”,”.”],[“Y”,”.”,”.”]]
Output: 3
Explanation:

Example 2:

Input: grid = [[“X”,”X”],[“X”,”Y”]]
Output: 0
Explanation:
No submatrix has an equal frequency of ‘X’ and ‘Y’.

Example 3:

Input: grid = [[“.”,”.”],[“.”,”.”]]
Output: 0
Explanation:
No submatrix has at least one ‘X’.

Constraints:

1 <= grid.length, grid[i].length <= 1000
grid[i][j] is either 'X', 'Y', or '.'.

Complexity Analysis

Time Complexity: O(mn)
Space Complexity: O(mn)

3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution in C++

class Solution {
 public:
  int numberOfSubmatrices(vector<vector<char>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    int ans = 0;
    // x[i][j] := the number of 'X' in grid[0..i)[0..j)
    vector<vector<int>> x(m + 1, vector<int>(n + 1));
    // y[i][j] := the number of 'Y' in grid[0..i)[0..j)
    vector<vector<int>> y(m + 1, vector<int>(n + 1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        x[i + 1][j + 1] =
            (grid[i][j] == 'X' ? 1 : 0) + x[i][j + 1] + x[i + 1][j] - x[i][j];
        y[i + 1][j + 1] =
            (grid[i][j] == 'Y' ? 1 : 0) + y[i][j + 1] + y[i + 1][j] - y[i][j];
        if (x[i + 1][j + 1] > 0 && x[i + 1][j + 1] == y[i + 1][j + 1])
          ++ans;
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution in Java

class Solution {
  public int numberOfSubmatrices(char[][] grid) {
    final int m = grid.length;
    final int n = grid[0].length;
    int ans = 0;
    // x[i][j] := the number of 'X' in grid[0..i)[0..j)
    int[][] x = new int[m + 1][n + 1];
    // y[i][j] := the number of 'Y' in grid[0..i)[0..j)
    int[][] y = new int[m + 1][n + 1];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        x[i + 1][j + 1] = (grid[i][j] == 'X' ? 1 : 0) + x[i][j + 1] + x[i + 1][j] - x[i][j];
        y[i + 1][j + 1] = (grid[i][j] == 'Y' ? 1 : 0) + y[i][j + 1] + y[i + 1][j] - y[i][j];
        if (x[i + 1][j + 1] > 0 && x[i + 1][j + 1] == y[i + 1][j + 1])
          ++ans;
      }

    return ans;
  }
}
// code provided by PROGIEZ

3212. Count Submatrices With Equal Frequency of X and Y LeetCode Solution in Python

class Solution:
  def numberOfSubmatrices(self, grid: list[list[str]]) -> int:
    m = len(grid)
    n = len(grid[0])
    ans = 0
    # x[i][j] := the number of 'X' in grid[0..i)[0..j)
    x = [[0] * (n + 1) for _ in range(m + 1)]
    # y[i][j] := the number of 'Y' in grid[0..i)[0..j)
    y = [[0] * (n + 1) for _ in range(m + 1)]

    for i, row in enumerate(grid):
      for j, cell in enumerate(row):
        x[i + 1][j + 1] = (cell == 'X') + x[i][j + 1] + x[i + 1][j] - x[i][j]
        y[i + 1][j + 1] = (cell == 'Y') + y[i][j + 1] + y[i + 1][j] - y[i][j]
        if x[i + 1][j + 1] > 0 and x[i + 1][j + 1] == y[i + 1][j + 1]:
          ans += 1

    return ans
 # code by PROGIEZ